Question

A clay ball of mass m and speed v strikes another metal ball of same mass m, which is at rest. They stick together after the collision. The kinetic energy of the system after collision is
A. $\dfrac{{m{v^2}}}{2}$
B. $\dfrac{{m{v^2}}}{4}$
C. $2m{v^2}$
D. $m{v^2}$

Answer 1

Here the velocities of clay ball and metal ball are given before collision. We have to find the velocity of the body after collision to calculate the kinetic energy. To calculate the velocity after collision, use the law of Conservation of Momentum.

Complete step by step answer:
We are given a clay ball of mass m and speed v strikes another metal ball of same mass m, which is at rest. They stick together after the collision.
We have to calculate the kinetic energy of the system after collision.
Kinetic energy (K.E) is the energy possessed by a body due to its motion.
$K.E = \dfrac{1}{2}m{v^2}$ , where m is the mass of the body and v is the velocity of the body.
To calculate the kinetic energy we need both mass and velocity.
The Law of Conservation of momentum states that for a collision occurring between two objects in an isolated system, the total momentum of the two objects before the collision is equal to the total momentum of the two objects after the collision.
Momentum= $m \times v$
Momentum before collision is ${m_1}{v_1} + {m_2}{v_2}$, where ${m_1}$ is the mass of clay ball and ${m_2}$ is the mass of metal ball; ${v_1}$ is the velocity of clay ball and ${v_2}$ is the velocity of the metal ball.
Velocity of the metal ball is 0 because it is at rest.
${m_1}{v_1} + {m_2}{v_2} \\\ {m_1} = m,{v_1} = v,{m_2} = m,{v_2} = 0 \\\ = {m_1}{v_1} + {m_2}{v_2} \\\ = m \times v + m \times 0 \\\ = mv \to eq(1) \\\$
Mass of the system after collision will be $m + m = 2m$, we have find the velocity
Momentum after collision is
$2m \times {v_f} \to eq(2) \\\ {v_f} = ? \\\$
Equate equations 1 and 2 as the momentums after and before collision must be the same.
$mv = 2m{v_f} \\\ {v_f} = \dfrac{{mv}}{{2m}} = \dfrac{1}{2}v \\\$
Kinetic energy after collision is
$K.E = \dfrac{1}{2}{m_f}{v_f}^2 \\\ \Rightarrow {m_f} = 2m,{v_f} = \dfrac{1}{2}v \\\ \Rightarrow K.E = \dfrac{1}{2} \times 2m \times {\left( {\dfrac{v}{2}} \right)^2} \\\ \Rightarrow K.E = \dfrac{1}{2} \times 2m \times \dfrac{{{v^2}}}{4} \\\ \Rightarrow K.E = \dfrac{{m{v^2}}}{4} \\\$
The kinetic energy of the system after collision is $\dfrac{{m{v^2}}}{4}$
Therefore, the answer is Option B, $\dfrac{{m{v^2}}}{4}$

Note: Kinetic energy of a system is the energy acquired by it by virtue of its motion whereas potential energy of a system is the energy acquired by it by virtue of its position. They are not the same. So do not confuse Kinetic Energy with Potential Energy.